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3URAL MATHEMATICAL JOURNAL, Vol. 7, No. 2, 2021, pp. 51-58
DOI: 10.15826/umj.2021.2.004
SHILLA GRAPHS WITH b = 5 AND b = 61
Alexander A. Makhnev^, Ivan N. Belousov^
Krasovskii Institute of Mathematics and Mechanics,
Ural Branch of the Russian Academy of Sciences, 16 S. Kovalevskaya Str., Ekaterinburg, 620990, Russia
Ural Federal University, 19 Mira str., Ekaterinburg, 620002, Russia
tmaldmev@imm.uran.ru, tti_belousov@mail.ru
Abstract: A Q-polynomial Shilla graph with b =5 has intersection arrays {105t, 4(21t + 1), 16(t + 1); 1, 4(t + 1), 84t}, t € {3, 4, 19}. The paper proves that distance-regular graphs with these intersection arrays do not exist. Moreover, feasible intersection arrays of Q-polynomial Shilla graphs with b = 6 are found.
Keywords: Shilla graph, Distance-regular graph, Q-polynomial graph.
1. Introduction
We consider undirected graphs without loops or multiple edges. For a vertex a of a graph r, denote by r(a) the ith neighborhood of a, i.e., the subgraph induced by r on the set of all vertices at distance i from a. Define [a] = r1(a) and a± = {a} U [a].
Let r be a graph, and let a, b € r. Denote by ^(a, b) (by A(a, b)) the number of vertices in [a] n [b] if a and b are at distance 2 (are adjacent) in r. Further, the induced [a] n [b] subgraph is called subgraph (A-subgraph).
If vertices u and w are at distance i in r, then we denote by bj(u, w) (by Cj(u, w)) the number of vertices in the intersection of Ti+1(u) (of ri-1(u), respectively) with [w]. A graph r of diameter d is called distance-regular with intersection array {b0, b1,..., bd-1; c1,..., cd} if, for each i = 0,..., d, the values b»(u, w) and c (u, w) are independent of the choice of vertices u and w at distance i in r. Define a» = k — b — c». Note that, for a distance regular graph, b0 is the degree of the graph and a1 is the degree of the local subgraph (the neighborhood of the vertex). Further, for vertices x and y at distance l in the graph r, denote by pj (x,y) the number of vertices in the subgraph r»(x) n Tj(y). The numbers pj(x,y) are called the intersection numbers of r (see [2]). In a distance-regular graph, they are independent of the choice of x and y.
A Shilla graph is a distance-regular graph r of diameter 3 with second eigenvalue equal to a = a3. In this case, a divides k and b is defined by b = b(r) = k/a. Morover, a1 = a — b and r has intersection array {ab, (a + 1)(b — 1), b2; 1, c2, a(b — 1)}. Feasible intersection arrays of Shilla graphs are found in [6] for b € {2, 3}.
Feasible intersection arrays of Shilla graphs are found in [1] for b = 4 (50 arrays) and for b = 5 (82 arrays). At present, a list of feasible intersection arrays of Shilla graphs for b = 6 is unknown. Moreover, the existence of Q-polynomial Shilla graphs with b = 5 also is unknown.
In this paper, we find feasible intersection arrays of Q-polynomial Shilla graphs with b = 6 and prove that Q-polynomial Shilla graphs with b = 5 do not exist.
1This work was supported by RFBR and NSFC (project № 20-51-53013).
Theorem 1. A Q-polynomial Shilla graph with b = 6 has intersection array
(1) {42t, 5(7t + 1), 3(t + 3); 1,3(t + 3), 35t}, where t € {7,12,17,27, 57};
(2) {372, 315, 75;1,15, 310}, {744, 625,125;1, 25, 620} or {930, 780,150; 1, 30, 775};
(3) {312,265,48;1,24, 260}, {624,525,80;1, 40, 520}, {1794,1500, 200; 1,100,1495} or {5694,4750,600;1, 300, 4745}.
In view of Theorem 2 from [1], a Q-polynomial Shilla graph with b = 5 has intersection array {105t, 4(21t + 1), 16(t + 1); 1, 4(t + 1), 84t}, t € {3, 4,19}.
Theorem 2. Distance-regular graphs with intersection arrays {315,256,64; 1,16, 252} and {1995, 1600, 320; 1, 80, 1596} do not exist.
Theorem 3. Distance-regular graphs with intersection array {420,340,80; 1,20,336} do not exist.
2. Proof of Theorem 1
In this section, r is a Q-polynomial Shilla graph with b = 6. Then (a2 — 5a — 6)2 — 4(5b2 — a2) is the square of an integer. By [6, Lemma 8], we have
2a < C2b(b + 1) + b2 — b — 2;
therefore, a < 21c2 +14. It follows from the proof of Theorem 9 in [6] that either k < b3 — b = 6 ■ 35 or v < k(2b3 — b + 1) = 428k. By [6, Corollary 17 and Theorem 20], the number b2 + c2 divides b(b — 1)b2 and
—34 = —b2 + 2 < 6>3 < —b2(b + 3)/(3b + 1) < —18. Theorem 2 from [7] implies the following lemma.
Lemma 1. If b2 = c2, then r has an intersection arrays {42t, 5(7t + 1), 3(t + 3);1, 3(t + 3), 35t} and t € {7,12,17,27, 57}.
To the end of this section, assume that b2 = c2 and k > > 02 > 03 are eigenvalues of the graph r. Then
6(6b2 + C2)/(b2 + C2) = —03.
On the other hand, according to [6, Lemma 10], the number c2 divides (a + 6)b2, 30a(a + 1) and (a + 6)b2 > (a + 1)c2.
Lemma 2. If —34 < 03 < —18, then one of the following statements holds:
(1) 03 = —31 and r has one of the intersection arrays {372,315,75; 1,15,310}, {744, 625,125; 1, 25, 620}, and {930, 780,150;1, 30, 775};
(2) 03 = —26 and r has one of the intersection arrays {312,265,48; 1,24,260}, {624, 525, 80; 1, 40, 520}, {1794,1500, 200; 1,100,1495}, and {5694, 4750, 600; 1, 300, 4745};
(3) 03 = —21 and r has one of the intersection arrays {42t, 5(7t + 1), 3(t + 3);1,3(t + 3), 35t} for t € {7,12,17, 27, 57}.
Proof. By [6, Lemma 10], c2 divides 6(6 — 1)62 = 3062 and, by [6, Corollary 17], the smallest nonprinciple eigenvalue 0з is equal to 6(662 + c2)/(62 + c2). Therefore, 30(0з + 6)/(0з + 36) is an integer and 6>з € {—34, —33, —32, —31, —30, —27, —26, —24, —21, —18}.
Let 03 = —34. Then 3(662 + c2) = 17(62 + c2) and 62 = 14c2. Further, 0з is a root of the equation x2 — (ai + a2 — k)x + (6 — 1)62 — a2 = 0; therefore, a = 425/28 ■ c2 — 34. In this case, the multiplicity of the first nonprincipal eigenvalue is m1 = 6/5 ■ (2545c2 — 5544)/c2, a contradiction with the fact that 5 does not divide 6 ■ 5544.
Let 6>з = —33. Then 2(662 + C2) = 11(6*2 + C2) and 62 = 9c2. Further, a = 275/27 ■ C2 — 33 and the multiplicity of the first nonprincipal eigenvalue is equal to m1 = 6/5 ■ (1645c2 — 5184)/c2, a contradiction as above.
Let 6>з = —32. Then 3(662 + C2) = 16(62 + C2) and 262 = 13c2. Further, a = 100/13 ■ C2 — 32 and the multiplicity of the first nonprincipal eigenvalue is m1 = 6/5 ■ (1195c2 — 4836)/c2, a contradiction as above.
Let 6>з = —31. Then 6(662 + C2) = 31(62 + C2) and 62 = 5c2. Further, a = 31/5 ■ C2 — 31 and the multiplicity of the first nonprincipal eigenvalue is m1 = 30(37c2 — 180)/c2 = 1110 — 5400/c2. The number of vertices in the graph is 31/5 ■ (222c2 — 2005c2 + 4500)/c2 ; hence, c2 divides 900 and is a multiple of 5. By computer enumeration, we find that, only for c2 = 15,25 and 30, we have admissible intersection arrays {372,315, 75; 1,15,310}, {744,625,125; 1,25,620} and {930,780,150; 1, 30, 775}.
Let 6>з = —30. Then (662 + C2) = 5(62 + C2) and 62 = 4c2. Further, a = 125/24 ■ C2 — 30 and the multiplicity of the first nonprincipal eigenvalue is m1 =6/5 ■ (745c2 — 4176)/c2, a contradiction as above.
Let 6>з = —27. Then 2(662 + C2) = 9(62 + C2) and 362 = 7c2. Further, a = 25/7 ■ C2 — 25 and the multiplicity of the first nonprincipal eigenvalue is m1 =6/5 ■ (445c2 — 3276)/c2, a contradiction as above.
Let 6>з = —26. Then 3(662 + C2) = 13(62 + C2) and 62 = 2c2. Further, a = 13/4 ■ C2 — 26 and the multiplicity of the first nonprincipal eigenvalue is m1 = 6(77c2 — 600)/c2 = 462 — 3600/c2. The number of vertices in the graph is 13/8 ■ (231c2 — 3340c2 + 12000)/c2; hence, c2 divides 1200 and is a multiple of 4. By computer enumeration, we find that only for c2 = 24,40,100, and 300 we have admissible intersection arrays {312,265,48; 1,24,260}, {624, 525,80; 1,40, 520}, {1794,1500,200;1,100,1495}, and {5694,4750,600; 1, 300,4745}.
Let 0з = —21. Then 2(662 + c2) = 7(62 + c2) and 62 = c2. Further, a = 7/3 ■ c2 — 21 and the multiplicity of the first nonprincipal eigenvalue is m1 = 6(41c2 — 360)/c2 = 246 — 2160/c2. The number of vertices in the graph is 7/3 ■ (82c2 — 1335c2 + 5400)/c2 ; hence, c2 divides 1080 and is a multiple of 3. By computer enumeration, we find that, only for c2 = 18,30,45,60,90, and 180, we have admissible intersection arrays {42t, 5(7t+1), 3(t+3); 1,3(t+3), 35t} for t € {3, 7,12,17,27, 57}. A graph with the array obtained for t = 3 does not exist by [5].
Let 6>з = —18. Then 6(662 + C2) = 19(62 + C2), so 362 = 2c2. Further, a = 2512 ■ C2 — 18 and the multiplicity of the first nonprincipal eigenvalue is m1 = 6/5 ■ (145c2 — 1224)/c2, a contradiction. The lemma is proved. □
Theorem 1 follows from Lemmas 1-2.
3. Triple intersection numbers
In the proof of Theorem 3, the triple intersection numbers [3] are used.
Let r be a distance-regular graph of diameter d. If u^u2,u3 are vertices of the graph r, then rj., r2, r3 are non-negative integers not greater than d. Denote by \ U^U3 ' the set of vertices
w € r such that d(w, u) = r and by
n\U2U'i
rir2r3
the number of vertices in
«1«2 «3 rir2r3
The numbers
«1«2«3 rir2r3
of
«1 «2«3 rir2r3
[rir2r3]. H
are called the triple intersection numbers. For a fixed triple of vertices ui,u2,u3, instead
, we will write [rir2r3]. Unfortunately, there are no general formulas for the numbers owever, [3] outlines a method for calculating some numbers [rir2r3]. Let u, v,w be vertices of the graph r, W = d(u, v), U = d(v,w), and let V = d(u, w). Since there is exactly one vertex x = u such that d(x,u) = 0, then the number [0jh] is 0 or 1. Hence [0jh] = ¿jw¿hV. Similarly, [i0h] = ¿¿w¿hU and [ij0] = ¿¿u ¿jV.
Another set of equations can be obtained by fixing the distance between two vertices from {u, v,w} and counting the number of vertices located at all possible distances from the third:
5>'h]= j - [0jh] i d
J>h] = - [iOh]
i d
E[iji]= - [ijo]
(3.1)
However, some triplets disappear. For |i — j| > W or i + j < W, we have pW = 0; therefore, [ijh] = 0 for all h € {0, ...,d}. We set
v, w) = ^ QriQsjQth
r,s,t=0
uvw rst -
If the Krein parameter qij = 0, then Sijh(u, v,w) = 0.
We fix vertices u,v,w of a distance-regular graph r of diameter 3 and set
{ijh} =
uvw
ijh
Calculating the numbers
[ijh] =
uvw ijh , [ijh]' = uwv ihj , [ijh]* = vuw _ jih J ' r. -7 wvu [ijh] = hji
uwv ihj , [ijh]* = vuw jih , [ijhr = wvu hji
[ijh]' =
(symmetrization of the triple intersection numbers) can give new relations that make it possible to prove the nonexistence of a graph.
d
4. Graphs with intersection arrays {315, 256, 64; 1,16, 252} and
{1995,1600,320;1,80,1596}
Let r be a distance-regular graph with intersection array {315,256,64; 1,16,252}. By [2, Theorem 4.4.3], the eigenvalues of the local subgraph of the graph r are contained in the interval [—5, 59/5). Since the Terwilliger polynomial (see [4]) is —4(5x — 59)(x + 5)(x + 1)(x — 43), then these eigenvalues lie in [—5, —1] U (59/5.43]. Hence, all nonprinciple eigenvalues are negative and the
local subgraph is a union of isolated (ai + 1)-cliques, a contradiction with the fact that a1 + 1 = 49 does not divide k = 315.
Thus, a distance-regular graph with intersection array {315,256,64; 1,16,252} does not exist.
Let r be a distance-regular graph with intersection array {1995,1600,320; 1,80,1596}. Then r has 1 + 1995 + 39900 + 8000 = 49896 vertices, spectrum 19951, 399495, 1523275, -2126125, and the dual matrix of eigenvalues
26125 -275
55. -209
The Terwilliger polynomial of the graph r is —20(x + 5)(x + 1)(x — 79)(x — 299); hence, the eigenvalues of the local subgraph are contained in [—5, —1] U {79} U {394}.
Note that the multiplicity m1 = 495 of the eigenvalue = 399 is less than k. By the corollary to Theorem 4.4.4 from [2] for b = 61/(01 + 1) = 4, the graph £ = [u] has an eigenvalue —1 — b = —5 of multiplicity at least k — m1 = 1500.
Let the number of eigenvalues 79 of the graph £ be equal to y. Then the sum of eigenvalues of the graph £ is at most —7500 — (494 — y) + 79y + 394; therefore, y > 95. Now twice the number of edges in £ is equal to
786030 = 1995 ■ 394 = ^ mi6>2
i
but not less than
25 ■ 1500 + 399 + 95 ■ 792 + 3942 = 786030.
Hence, £ has spectrum 3941.7995, —1399, — 51500.
Now the number t = ksAs/2 of triangles in £ containing this vertex is equal to ^i mi03/(2v). Therefore,
t = Y1 mA3/(2v) = (3943 + 793 ■ 95 — 399 — 125 ■ 1500)/3990 = 27021
i
and As = 54042/394 is approximately equal to 137.16, a contradiction.
Thus, a distance-regular graph with intersection array {1995,1600,320; 1,80,1596} does not exist.
Theorem 2 is proved.
5. Graph with array {420, 340, 80; 1, 20, 336}
Let r be a distance-regular graph with intersection array {420,340,80; 1,20,336}. Then r is a formally self-dual graph having 1 + 420 + 7140 + 1700 = 9261 vertices, spectrum 4201, 84420 , 07140, —211700, and the dual matrix of eigenvalues
1700 \
—85 .
20 . —64 J
The Terwilliger polynomial of the graph r is —20(x+5)(x+1)(x—16)(x —59) and the eigenvalues of the local subgraph are contained in [—5, —1] U {16} U {79}. If the nonprinciple eigenvalues of a local subgraph are negative, then this subgraph is a union of isolated (a1 + 1)-cliques, a contradiction with the fact that a1 + 1 = 80 does not divide k = 420. Hence, the local subgraph has eigenvalue 6.
Q =
1 495 23275
1 99 175
1 0 —56
1 —99/4 931/4
Q =
1 420 7140
1 84 0
1 0 —21
1 —21 84
Lemma 3. Intersection numbers of a graph r satisfy the equalities
(1) pii = 79, p2i = 340, p32 = 1360, P22 = 5440, p33 = 340,
(2) p2i = 20, p22 = 320, p23 = 80, p22 = 5519, p23 = 1300, p23 = 320;
(3) p?2 = 336, p?3 = 84, p22 = 5460, p33 = 1344, p33 = 271.
Proof. Direct calculations. □
Let u, v, and w be vertices of a graph r, [rst] = [UVT], ^ = r3(u), and let E = Q2. Then E is a regular graph of degree 1344 on 1700 vertices.
Lemma 4. Let d(u, v) = d(u, w) = 3 and d(v,w) = 1. Then the following equalities hold:
(1) [122] = 2re/5 — 136, [123] = [132] = —2re/5 + 472, [133] = 2re/5 — 388;
(2) [211] = re/10 — 38, [212] = [221] = —re/10 + 374, [222] = —14r6/10 + 5576, [223] = [232] = 3re/2 — 490, [233] = —3re/2 + 1834;
(3) [311] = —re/10 + 117, [312] = [321]=re/10 — 34, [322] = re, [323] = [332] = —11re/10 + 1378, [333] = 11re/10 — 1107,
where re € {1010,1020,... , 1170}.
Proof. A simplification of formulas (3.1) taking into account the equalities Sii3(u,v,w) = Si3i(u, v,w) = S3ii(u, v,w) =0. □
By Lemma 4, we have 1010 < [322] = re < 1170.
Lemma 5. Let d(u, v) = d(u, w) = d(v,w) = 3. Then the following equalities hold:
(1) [122] = — ri7 + 336, [123] = [132] = ri7, [133] = — ri7 + 84;
(2) [213] = [231] = ri7, [212] = [221] = — ri7 + 336, [222] = 39ri7/4 + 3444, [223] = [232] = —35ri7/4 + 1680, [233] = 31ri7/4 — 336;
(3) [313] = [331] = — ri7 + 84, [312] = [321] = ri7, [322] = —35ri7/4 + 1680, [323] = [332] = 31ri7/4 — 336, [333] = —27ri7/4 + 522,
where ri7 € {44, 48,... , 76}.
Proof. A simplification of formulas (3.1) taking into account the equalities Sii3(u,v,w) = Si3i(u, v,w) = S3ii(u, v,w) =0. □
By Lemma 5, we have 1015 < [322] = —35ri7/4 + 1680 < 1295.
The number d of edges between E(w) and E — ({w} U A(w)) satisfies the inequalities
359905 = 84 ■ 1010 + 271 ■ 1015 < d < 84 ■ 1170 + 271 ■ 1295 = 449225, 267.786 < 1343 — A < 334.245, 1008.755 < A < 1075.214,
where A is the mean value of the parameter A(E).
Lemma 6. Let d(u, v) = d(u, w) = 3 and d(v,w) = 2. Then the following equalities hold:
(1) [122] = (—64r15 + 4r16 + 7364)/27, [123] = [132] = (64r15 — 4r16 + 1708)/27, [133] = (—64r15 + 4r16 + 560)/27;
(2) [211] = — r15+20, [212] = [221] = (71r15+4r16+6392)/27, [222] = (—17r15 — 13r16+38311)/9, [223] = [232] = (—20r15 + 35r16 + 26095)/27, [233] = (64r15 — 31r16 + 8053)/27;
(3) [311] = r15, [312] = [321] = (—71r15 — 4r16 + 2248)/27, [313] = (44r15 + 4r16 + 20)/27, [322] = (115r15 + 35r16 + 26716)/27, [323] = [332] = (—44r15 — 31r16 + 7297)/27, [333] = r16,
where — 10r15 + 4r16 + 20 is a multiple of 27, r15 € {0,1,..., 20}, and r16 € {0,1,..., 235}.
Proof. A simplification of formulas (3.1) taking into account the equalities S113(u,v,w) = S131(u, v,w) = S311(u, v,w) =0. □
By Lemma 6, we have
998 < [322] = (115r15 + 35r16 + 26716)/27 < 1294. Let us count the number h of pairs of vertices y and z at distance 3 in the graph Q, where
On the one hand, by Lemma 4, we have [323] = -11r6/10 + 1378, where r6 € {1010,1020,..., 1170},
7644 = 8491 < h < 84267 = 22428.
On the other hand, by Lemma 6, we have [313] = (44r15 + 4r16 + 20)/27, where r15 € {0,1,..., 20}, r16 € {0,1,..., 235}, therefore
7644 < Y^(44ri5 + 4ri6) + 995.55 < 22428,
therefore
6648.44 < Y^(44ri5 + 4ri6) < 21432.45,
i
4.946 < Y(11rl5 + ri6)/1344 < 15.947.
If ri 5 = 0, then ri 6 + 5 is a multiple of 27 and ri6 = 22.49,.... If ri 5 = 1, then 2ri 6 + 5 is a multiple of 27 and ri 6 = 11.38,.... In any case,
E(11ri5 + ri6)/1344 > 22,
a contradiction.
Theorem 3 is proved.
□
Conclusion
The following are the main steps in creating a theory of Shilla graphs:
(1) finding a list of feasible intersection arrays of Shilla graphs with b = 6;
(2) classification of Q-polynomial Shilla graphs with b2 = c2.
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